【杭电5835】Danganronpa

Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 717    Accepted Submission(s): 510

Problem Description

Chisa Yukizome works as a teacher in the school. She prepares many gifts, which consist ofn kinds with a[i] quantities of each kind, for her students and wants to hold a class meeting. Because of the busy work, she gives her gifts to the monitor, Chiaki Nanami. Due to the strange design of the school, the students' desks are in a row. Chiaki Nanami wants to arrange gifts like this:

1. Each table will be prepared for a mysterious gift and an ordinary gift.

2. In order to reflect the Chisa Yukizome's generosity, the kinds of the ordinary gift on the adjacent table must be different.

3. There are no limits for the mysterious gift.

4. The gift must be placed continuously.

She wants to know how many students can get gifts in accordance with her idea at most (Suppose the number of students are infinite). As the most important people of her, you are easy to solve it, aren't you?

 

Input

The first line of input contains an integer T(T10) indicating the number of test cases.

Each case contains one integer n. The next line contains n(1n10) numbers: a1,a2,...,an,(1ai100000).

 

Output

For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the answer of Chiaki Nanami's question.

 

Sample Input

123 2

 

Sample Output

Case #1: 2

 

Author

UESTC

 

Source

2016中国大学生程序设计竞赛 - 网络选拔赛

题意：有n种礼物，分给若干个同学，每个同学至少有一个神秘礼物和一个普通礼物，相邻同学的普通礼物不能相同，神秘礼物没有要求，在此条件下最多可以分给多少个同学。因题中没有特别要求，所以一种礼物既可以做普通礼物也可以做神秘礼物，每人至少两个礼物，最多分给sum/2个同学

#include<cstdio>int main(){int t,n,k=1;int a[100];scanf("%d",&t);while(t--){int sum=0;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum+=a[i];}printf("Case #%d: %d\n",k++,sum/2);}return 0;}

如果按上面的思路，那么假如只有一种礼物，有五个，可以分给两个人，但是由于连着发，他们的普通礼物就相同了，显然不对。

另一种思路：

sum/2是最多的情况。什么时候发不了那么多，就是当max很大，无论怎么发，都发不完max。那这时候贪心策略就是一个小朋友发max*2，下一个发max+其它的，再下一个发max*2，再下一个发max+其它的...。

所以答案是sum/2和（sum-max）*2（一个其它的和3个数量最多的那种礼物组合发给两个小朋友）+1（多出的amax发给一个小朋友）的最小值。

#include <cstdio>int t,n,x,sum,m;int main() {    scanf("%d",&t);    for(int cas=1;cas<=t;cas++){        sum=m=0;        scanf("%d",&n);        for(int i=1;i<=n;i++){            scanf("%d",&x);            sum+=x;            m=max(m,x);        }        printf("Case #%d: %d\n",cas,min(sum/2,(sum-m)*2+1));    }}

两个代码都可以通过，感觉下面的更准确吧